TUTORIAL #4-2014
1
T UTORIAL 4
Question 1
Let vp denote a real passband signal, with Fourier transform Vp (f ) specified as follows for negative
frequencies:
Vp (f ) =


f + 101,

0,
−101 ≤ f ≤ −99
(1)
f < −101 or − 99 < f < 0.
(a) Sketch Vp (f ) for both positive and negative frequencies.
(b) Without explicitly taking the inverse Fourier transform, can you say whether vp (t) = vp (−t) or
not?
Solution:
(a) From Eq.(1), the passband spectrum Vp (f ) is sketched in Fig.1, where the Fourier transform property
that Vp (f ) = Vp∗ (−f ), since vp (t) is real-valued.
Fig. 1.
Passband spectrum for 1) as functions of f
(b) From Fig.1, since Vp (f ) = Vp (−f ) then vp (t) = vp (−t) using the even property of Fourier
Transform.
Question 2:
Consider the following two passband signals:
√
up (t) = 2sinc(2t) cos 100πt
and
vp (t) =
February 9, 2014
√
π
2sinc(t) sin 101πt +
.
4
Dr. J. Chen, EE 4TM4: Digital Communications II
TUTORIAL #4-2014
2
(a) Find the complex envelopes u(t) and v(t) for up and vp , respectively, with respect to the frequency
reference fc = 50.
(b) What is the bandwidth of up (t)? What is the bandwidth of vp (t)?
(c) Find the inner product hup , vp i, using the result in (a).
(d) Find the convolution yp (t) = (up ∗ vp )(t), using the result in (a).
Solution:
(a) For a frequency reference fc = 50, clearly the complex envelope for up is given by u(t) = sinc(2t).
For vp , we write
√
π
2sinc(t) sin 101πt +
4
√
j(101πt+ π4 )
=
2Re sinc(t) −je
√
π
=
2Re sinc(t)ej(πt− 4 ) ej100πt .
vp (t) =
(2)
Thus, from Eq.(2),
π
v(t) = sinc(t)ej(πt− 4 ) .
(3)
u(t) = sinc(2t).
(4)
Similarly
(b) From Eq.(3), note that U (f ) occupies a band of length 2, while V (f ) occupies a band of length
1. Thus, up has a bandwidth of 2, and vp has a bandwidth of 1.
(c) The inner product
Z
hup , vp i = Re (hu, vi) = Re
∗
U (f )V (f )df
,
π
but U (f ) = 21 I[−1,1] (f ) and V (f ) = e−j 4 I[0,1] (f ), so that
Z
1 π
U (f )V ∗ (f )df = ej 4 .
2
Therefore, we obtain that
1
hup , vp i = √ .
2 2
February 9, 2014
(5)
Dr. J. Chen, EE 4TM4: Digital Communications II
TUTORIAL #4-2014
(d) We have Y (f ) =
3
√1 U (f )V
2
1
√
V
2 2
(f ) (since U (f ) takes the constant value
1
1
π
support of V (f )). This implies that yp (t) = 2√
v
(t)
=
sinc(t)
sin
101πt
+
.
p
2
4
2
February 9, 2014
(f ) =
1
2
over the
Dr. J. Chen, EE 4TM4: Digital Communications II